Given, ΔABC is an equilateral triangle.
Here, AB = 8 units
∴ AB = BC = CA = 8 units
Thus,ΔADE is an equilateral triangle.
Here, E is the mid-point of AB
\(\therefore AE=BE=\frac { AB }{ 2 } =\frac { 8 }{ 2 } =4units\)
Now, in ΔADE, AD = DE = EA = 4 units
Similarly, equilateral triangles are ΔBOT and ΔUPC
having each sides equal i.e.
BO = OT = BT = UC = PC = PU = 4 units
It is also clear that, OC = PA = 4 units
Also, ΔDIF is an equilateral triangle.
Here, F is the mid-point of DE.

\(\therefore DF=FE=\frac { DE }{ 2 } =\frac { 4 }{ 2 } =2units\)
In ΔDIF, DI = IF = DF = 2 units
Similarly, in ΔTKN and ΔRQU,
TK = KN = TN = RQ = UQ = UR = 2 units
It is also clear that NO = RP = 2 units
Also, ΔHIG is an equilateral triangle.
Here, G is the mid-point of IF
\(\therefore IG=GF=\frac { IF }{ 2 } =\frac { 2 }{ 2 } =1\)
Now, in ΔHIG, HG = HI = GI = 1 unit
Similarly, in ΔMLK and ΔXQS,
ML = MK = LK = SQ = XS = QX = 1unit
It is also clear that, LN = XR = 1 unit
Now, perimeter of the given figure
= Sum of all outer sides of the given figure
= AD + DI + IH + HG + GF + FE + EB + BT +TK + KM + LM + LN+NO+OC +CU+UO + OS + XS + XR + PR +PA
=[4+ 2+1 +1 +1 + 2+ 4+ 4+ 2+1 +1+1+2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4] cm
=45cm
Hence, the perimeter of the given figure is 45 cm.

Given numbers can be represented in decimals as follows:
(i) 2 \(\times\) hundreds + 1 \(\times\) ten + 6 \(\times\) ones + 3 \(\times\) tenths
= 2 \(\times\) 100 + 1 \(\times\) 10 + 6 \(\times\) 1+ 3 \(\times\) \(\frac { 1 }{ 10 } \)
= 200 + 10 + 6 + \(\frac { 3 }{ 10 } \) = 216 + 0.3 = 216.3
(ii) 4 \(\times\) hundreds + 5 \(\times\) tens + 4 \(\times\) ones + 2 \(\times\) tenths
= 4 \(\times\) 100 + 5 \(\times\) 10 + 4 \(\times\) 1+ 2 \(\times\) \(\frac { 1 }{ 10 } \)
= 400 + 50 + 4 + \(\frac { 2 }{ 10 } \) = 454+0.2= 454.2
(iii) 7 \(\times\)hundreds + 3 \(\times\) tens + 2 \(\times\) ones + 1 \(\times\) tenth
 = 7 \(\times\) 100 + 3 \(\times\) 10 + 2 \(\times\) 1+ 1 \(\times\) \(\frac { 1 }{ 10 } \)
= 700 + 30 + 2 + \(\frac { 1 }{ 10 } \) = 732 + 0.1 = 732.1

Let width of rectangle (b)= x cm
Then, length of rectangle (I) = 3x cm
∴ Perimeter = 2 (l + b)
⇒ 40 =2 (3x + x)
⇒ 8x =40
⇒ x=\(\frac{40}{8}\)=5cm
Hence, length is 15 cm and width is 5 cm.

Required area is 245 sq cm

It represents 5 (frequency).

Let flying time of bird be t min.
Then, bird flies in one min = 1 km
∴ Bird flies in t min = t\(\times\)t km = t km
Hence, the distance covered by the bird in its flying time i.e. in t min is t km.

We know that, 10 mm = 1 cm \(\Rightarrow\) 1 mm = \(\frac {1}{10}\) cm
\(\therefore \quad 30\quad mm=30\times \frac { 1 }{ 10 } cm=\frac { 30 }{ 10 } cm=3\quad cm\)

\(19\frac { 3 }{ 100 } =19+\frac { 3 }{ 100 } \)= 19 + 0.03 = 19.03

By inspection of the above table, we find that t =7
satisfies the equation 5t = 35. [∵ at = 7, LHS = RHS]
Hence, t = 7 is its solution.

(a) 25
(b) 20
(c) 15
(d) A
(e) 55.

r

Let ABCD be the page of a notebook.
On measuring, AB = CD = 15 cm and BC = DA = 20 cm

Then, the sum of the lengths of four sides
= AB + BC + CD + DA
= 15 cm + 20 cm + 15 cm + 20 cm
= (I5 + 20 + 15 + 20) cm = 70 cm
Hence, the sum of the four sides is 70 em.
Now, perimeter of the page = Sum of the lengths of four sides
=70 cm